Chemistry Set I

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Physics Set 2

Physics Set 1

Class 10 Opt Science Question

Diffraction

aDiffractionThe phenomenon of bending of light from the corners or edges of aperture or obstacles and departure of true rectilinear path of light is diffraction of light. It is due to the super position secondary wavelets generated from single wave front.Types of Diffraction 1. Fresnel diffraction: If the source of light and screen is finite distance from the slit, then this type of diffraction is Fresnel diffraction. No lenses are needed to converge the light at slit and screen. The incident wave front on slit is spherical or cylindrical. 2. Fraunhofer diffraction: If the source of light and screen is infinite distance from the slit, then this type of diffraction is Fraunhofer diffraction. Two lenses are used to converge the light at slit and screen. The incident wave front on slit is plane wave front. Diffraction through Single slit ( Fraunhofer diffraction)

Consider a plane wavefront is incident on a narrow slit a shown in fig. Suppose a convex lens of focallength 'f' is placed just behind the slit such that f≈D$f\approx D$, distance between the sit and the screen. Here each point on the wavefront act as a source of light in same phase. When the light from these points reach on the screen, they interfere in different conditions. The light reaching at the centre of the screen reaches in the same phase and bright fringe is obtained called central maxima. Then on both sides of the centre of the screen, dark and bright fringes are formed called secondary minima and secondary maxima respectively.Condition for secondary minima:Here, when n=1$n=1$, divide the slit into two equal parts. Now, the pair of light going from upper and lower part will have path difference 𝜆

2$\frac{𝜆}{2}$ and 1st minima is formed. When n=2$n=2$, divide the slit into four equal parts. Now, light going from upper and consecutive lower part will have path difference 𝜆

2$\frac{𝜆}{2}$ and 2nd minima is formed, and so on. Here, secondary maxima will be formed between secondary minima. For minima , path difference, S1N=n𝜆${\mathrm{S}}_{1}N=n𝜆$.... (i), n=1,2,…$n=1,2,\dots$In 𝛥S1S2N,sin𝜃n=S2N

S1S2$𝛥{\mathrm{S}}_1{\mathrm{S}}_{2}N,\sin {𝜃}_n=\frac{{\mathrm{S}}_{2}N}{{\mathrm{S}}_1{\mathrm{S}}_2}$of, S2N=S1S2sin𝜃n ${\mathrm{S}}_{2}N={\mathrm{S}}_1{\mathrm{S}}_2\sin {𝜃}_n$of. S2N=asin𝜃n…${\mathrm{S}}_{2}N=a\sin {𝜃}_{\mathrm{n}}\dots$ (ii)From (i) and (ii) a sin𝜃n=n𝜆$\sin {𝜃}_n=n𝜆$ or, sin𝜃n=n𝜆

a$\sin {𝜃}_n=\frac{n𝜆}{a}$...(iii)If yn$y_n$ be the distance of nth $n^{\text{th }}$ minima from the centre of the screen,tan𝜃n= yn

D................ (iv) $$\tan {𝜃}_n=\frac{y_n}{D}\text{................ (iv) }$$Here, 𝜃n${𝜃}_n$ is very small∴ tan𝜃n≈sin𝜃n$\therefore \tan {𝜃}_n\approx \sin {𝜃}_n$or. yn

D=n␒𝜆

a$\frac{y_n}{D}=\frac{n\stackrel{.}{𝜆}}{a}$or, yn=n𝜆D

a $y_n=\frac{n𝜆D}{a}$This is the distance of nth $n^{\text{th }}$ minima from the centre of the screen. ∴ $\therefore$ Width of secondary minima, 𝛽=yn+1-yn$𝛽=y_{n+1}-y_n$or, 𝛽=(n+1)𝜆D

d-n𝜆D

d$𝛽=\frac{(n+1)𝜆D}{d}-\frac{n𝜆D}{d}$or, 𝛽=n𝜆D

d+𝜆D

d-n𝜆D

d$𝛽=\frac{n𝜆D}{d}+\frac{𝜆D}{d}-\frac{n𝜆D}{d}$or, 𝛽=𝜆⋅D

d$𝛽=\frac{𝜆\cdot D}{d}$ Condition for secondary maxima:Here, when n=1$n=1$, divide the slit into three equal parts. The light reaching at p$p$ from pair of points from two consecutive parts will be out of phase and the light from the remaining one part produces 1 st secondary maxima. When n=2$n=2$, divide the slits into 5 equal parts. Here, pair of light coming from consecutive parts of the slit will be out of phase and the light from the remaining part produces 2nd $2^{\text{nd }}$ secondary maxima and so on.For maxima, path difference S2N= (2n+1)𝜆

2,n=1,2,…$S_{2}N=(2n+1)\frac{𝜆}{2},n=1,2,\dots$.or, asin𝜃n=(2n+1)𝜆

2$a\sin {𝜃}_n=(2n+1)\frac{𝜆}{2}$or, sin𝜃n=(2n+1)𝜆

2a…...........(v)${\displaystyle \sin }{𝜃}_n={\displaystyle \frac{(2n+1)𝜆}{2a}}\dots ...........(v)$If yn$y_n$ be the distance of nth $n^{\text{th }}$ maxima from the centre of the screen,tan𝜃n=Yn

D… (vi) $$\tan {𝜃}_n=\frac{Y_n}{D}\dots \text{ (vi) }$$Here, 𝜃n${𝜃}_n$ is very small.∴ tan𝜃n=sin𝜃n$\therefore \tan {𝜃}_n=\sin {𝜃}_n$or, yn

D=(2n+1)𝜆

2d$\frac{y_n}{D}=\frac{(2n+1)𝜆}{2d}$or, yn=(2n+1)𝜆D

2a$y_n=\frac{(2n+1)𝜆D}{2a}$This is the distance of nth $n^{\text{th }}$ secondary maxima from the centre of the screen. ∴$\therefore$ Width of secondary maxima,𝛽′=yn-yn-1$${𝛽}^{\prime }=y_n-y_{n-1}$$or, 𝛽=(2n+1)𝜆D

2a-[2(n-1)+1]𝜆D

2a$𝛽=\frac{(2n+1)𝜆D}{2a}-\frac{[2(n-1)+1]𝜆D}{2a}$or, 𝛽′=(2n+1)𝜆D

2a-(2n+1)𝜆D

2a+2𝜆D

2a${𝛽}^{\prime }=\frac{(2n+1)𝜆D}{2a}-\frac{(2n+1)𝜆D}{2a}+\frac{2𝜆D}{2a}$or, 𝛽′=𝜆D

a${𝛽}^{\prime }=\frac{𝜆D}{a}$∴$\therefore$ Width of secondary maxima = width of secondary minima.Central maxima: Central maxima occurs due to the superposition of secondary wavelets reaching at Q on the screen constructively, ie all the wave are in phase. It is formed between 1st minima on both sides of the screen from the centre of the screen.∴ $\therefore$ Width of central maxima, a

𝛽0	=y1+y1
or, 𝛽0	=1×𝜆D a+1×𝜆D a
or, 𝛽0	=2𝜆D a

$$\begin{array}{rl}{𝛽}_0& =y_1+y_1\\ \text{ or, }{𝛽}_0& =\frac{1\times 𝜆D}{a}+\frac{1\times 𝜆D}{a}\\ \text{ or, }{𝛽}_0& =\frac{2𝜆D}{a}\end{array}$$

Diffraction Grating

An arrangement consisting of a large number of narrow parallel slits of equal widih and are seperaled one another by the large number of opaques having equal widths is called diftraction grating. In diftraction grating, the large number of slits and opeques are drawn with the help of fine sharp diamond point. The seratch due to the diamond point is laken as opeque and the transparenly is taken as silt. If such arrangement is oblained in the glass plate, it is called tramission grating.The above figure shows the plane transmission grating Let the with of opaque portion be b$b$. The distance (a+b)$(a+b)$ is called grating spacing. Consider a parallel beam of monochromatic light having wave length 𝜆${\displaystyle 𝜆}$ incident normally on the grating surface, where most of light is brought to focus by the lens at point 0 gives rise to the central maxima and rest of the light spread in all the directions. Consider light waves making angle 𝜃$𝜃$ with the initial direction, then path difference between these waves reaching to the point P$P$ isPath difference =(a+b) sin𝜃$=(a+b)\sin 𝜃$ For the point P$P$ to be the point of maximum intensity. Path difference = n𝜆$Pathdifference=n𝜆$i.e. (a+b)sin𝜃=n𝜆$(a+b)\sin 𝜃=n𝜆$So in general,(a+b) sin𝜃n=n𝜆$\sin {𝜃}_n=n𝜆$If N$N$ be the number of lines per unit length then N=1

a+b$N=\frac{1}{a+b}$a

(a+b)	=1 N
∴ 1 N sin𝜃n	=n𝜆

$$\begin{array}{rl}(a+b)& =\frac{1}{N}\\ \therefore \frac{1}{N}\sin {𝜃}_n& =n𝜆\end{array}$$ Resolving PowerThe resolving power of an optical instrument is the ability of an instrument to produce distinctly separate images of two close objects. When the central maxima of one image falls on the first minima of another image, the image is said to be just resolved. This limiting condition of resolution is known as Rayleigh's criterion.Resolving power of microscopeConsider a point object O$O$ illuminated by a light of wavelength 𝜆$𝜆$ and it is observed through a microscope. If 𝜃$𝜃$ is the semi-vertical angle , the least distance between to objects which can be distinguished is given byd=𝜆

2𝜇sin𝜃$$d=\frac{𝜆}{2𝜇\sin 𝜃}$$where 𝜇$𝜇$ is the refractive index of the medium between the object and objective lens. The resolving power of the microscope is the reciprocal of the minimum separation between two objects between two objects seen distinctly.∴$\therefore$ Resolving power of microscope =1

d=2𝜇sin𝜃

𝜆$=\frac{1}{d}=\frac{2𝜇\sin 𝜃}{𝜆}$From above equation, we observe that the resolving power of microscope increases when the refractive index of the medium is increased.Resolving Power of a TelescopeIt is reciprocal of the smallest angular separation between two distinct objects whose images are separated in a telescope. The angular separation is given byd𝜃=1.22𝜆

D$$d𝜃=1.22\frac{𝜆}{D}$$where d𝜃$d𝜃$ is the angle subtended at the objective, 𝜆=$𝜆=$ wavelength of light used, and D$D$ is the diameter of telescope objective. Resolving power of telescope =1

d𝜃=1

1.22 D

𝜆$=\frac{1}{d𝜃}=\frac{1}{1.22}\frac{D}{𝜆}$. So resolving power of a telescope can be increased then by increasing the diameter of the objective or decreasing the wavelength of light.